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An RC circuit has a time constant of 3.1 s. At \(t=0\), the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

Short Answer

Expert verified
Answer: The energy stored in the capacitor reaches half of its maximum value at approximately \(1.08\) seconds.

Step by step solution

01

Express the half energy voltage in terms of maximum voltage

We are given the energy at half the maximum capacity, and we need to find the corresponding voltage. Using the energy formula, we have: \(E_\frac{1}{2} = \frac{1}{2}CV^2_\text{half} = \frac{1}{4}CV^2_\text{max}\) Dividing both sides by \(\frac{1}{2}C\): \(V^2_\text{half} = \frac{1}{2}V^2_\text{max}\) Taking a square root of both sides: \(V_\text{half} = V_\text{max}\sqrt{\frac{1}{2}}\)
02

Write the charging equation for the half energy voltage

Plugging in the half voltage value into the charging equation: \(V_\text{max}\sqrt{\frac{1}{2}} = V_0(1 - e^{-\frac{t}{RC}})\) Where our goal is to find the time \(t\).
03

Solve for t in the charging equation

Now we will solve for \(t\) in the equation from Step 2: \(\sqrt{\frac{1}{2}} = 1 - e^{-\frac{t}{RC}}\) Rearrange the equation to isolate the exponential term: \(e^{-\frac{t}{RC}} = 1 - \sqrt{\frac{1}{2}}\) Take the natural logarithm of both sides: \(-\frac{t}{RC} = \ln(1 - \sqrt{\frac{1}{2}})\) Multiply both sides by \(-RC\) to find \(t\): \(t = -RC\ln(1 - \sqrt{\frac{1}{2}})\)
04

Plug the given time constant into the equation for t

We are given the time constant \(RC = 3.1\,\text{s}\), so we plug this value into the equation to find \(t\): \(t = -3.1\,\text{s}\ln(1 - \sqrt{\frac{1}{2}})\) Evaluate the expression to find the time: \(t \approx 1.08\,\text{s}\) So, at approximately \(1.08\) seconds, the energy stored in the capacitor will reach half of its maximum value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitor Charging
When a capacitor is placed in an electric circuit, especially in an RC circuit, it begins the process of charging as soon as the circuit is completed. A capacitor stores electrical energy in the electric field between its plates, and the voltage across a charging capacitor can be described through an exponential equation.
Understanding this equation is crucial to determining the voltage at any given time during the charging process. The equation can be expressed as:
  • \( V(t) = V_0(1 - e^{-\frac{t}{RC}}) \), where \( V_0 \) is the maximum voltage the capacitor will reach once fully charged.
  • \( RC \) is known as the time constant, and it describes how quickly the capacitor charges.
During the initial phase of capacitor charging, the voltage rises quickly, but as it approaches its maximum value, the rate of change slows down. At any point, by this formula, you can calculate how much charge has accumulated on the capacitor.
Time Constant
The time constant, often represented by the symbol \( RC \), is a fundamental parameter in any RC circuit. It is the product of the resistance \( R \) and the capacitance \( C \) of the capacitor. This value, expressed in seconds, indicates how fast a capacitor in the circuit will charge or discharge.
Let's break it down:
  • After a time period equal to one time constant, \( 63.2\% \) of the maximum charge is reached during the charging process.
  • After five times the time constant, the capacitor is virtually completely charged, reaching over \( 99\% \) of its final charge.
In practical applications, smaller time constants result in faster charging capacitors, while larger values mean the capacitor charges more gradually. This can be important in applications where timing and smooth transitions are essential.
Electric Circuits
Electric circuits, a fundamental concept in physics and electrical engineering, are pathways that allow the flow of electric current. They can be simple, with just a power supply, a load, and connecting wires, or more complex, involving components like resistors, capacitors, and inductors.
Different circuit types include:
  • Series Circuits: Components are arranged in a single path, so the same current flows through all components.
  • Parallel Circuits: Each component has its own path to the voltage source, allowing current to flow through multiple paths.
In RC circuits, capacitors and resistors are arranged in series, and the behavior of the circuit can be determined by analyzing the voltage and current over time. Understanding these concepts is essential for designing circuits with specific time responses, such as in filters and signal processing.
Energy Storage
Energy storage in capacitors plays a critical role in many electronic applications. A capacitor stores energy in the form of an electric field between its plates and can release it quickly when needed. This ability to charge and discharge rapidly makes capacitors very useful as temporary energy storage devices in various electronic circuits.
The energy stored in a capacitor is given by the formula:
  • \( E = \frac{1}{2}CV^2 \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
Capacitors can provide a burst of power needed by circuits, helping to stabilize voltage and power flow in electronic systems. They are also used in applications like power conditioning, signal processing, and in maintaining power supply during brief outages. Understanding how capacitors store and release energy is crucial in designing systems for a wide array of applications.

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Most popular questions from this chapter

Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?

In the movie Back to the Future, time travel is made possible by a flux capacitor, which generates 1.21 GW of power. Assuming that a 1.00 - F capacitor is charged to its maximum capacity with a \(12.0-\mathrm{V}\) car battery and is discharged through a resistor, what resistance is necessary to produce a peak power output of 1.21 GW in the resistor? How long would it take for a \(12.0-\mathrm{V}\) car battery to charge the capacitor to \(90.0 \%\) of its maximum capacity through this resistor?

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?

The dead battery of your car provides a potential difference of \(9.950 \mathrm{~V}\) and has an internal resistance of \(1.100 \Omega\). You charge it by connecting it with jumper cables to the live battery of another car. The live battery provides a potential difference of \(12.00 \mathrm{~V}\) and has an internal resistance of \(0.0100 \Omega\), and the starter resistance is \(0.0700 \Omega\). a) Draw the circuit diagram for the connected batteries. b) Determine the current in the live battery, in the dead battery, and in the starter immediately after you closed the circuit.

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

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